By factoring the expression, check that n³ − n is always divisible by 6 for all natural numbers n. Give reasons.

Step-by-step solution

Idea: A challenge question. Factorise n³ − n using a² − b² = (a − b)(a + b). The factors turn out to be three numbers in a row, and among any three numbers in a row there is always an even number and always a multiple of 3.

1. Factorise. Take n common, then use a² − b² = (a − b)(a + b):
   n³ − n = n(n² − 1) = n(n − 1)(n + 1) = (n − 1) × n × (n + 1)1 mark
2. n − 1, n, n + 1 are three consecutive integers (numbers in a row, like 4, 5, 6).½ mark
3. Divisible by 2: in any two numbers in a row, one is even. So at least one of the three factors is even, and the product is divisible by 2.½ mark
4. Divisible by 3: every third number is a multiple of 3, so one of any three numbers in a row is a multiple of 3. So the product is divisible by 3.½ mark
5. 2 and 3 have no common factor other than 1, so a number divisible by both 2 and 3 is divisible by 2 × 3 = 6. Hence n³ − n is divisible by 6 for every natural number n. (For n = 1 the value is 0, and 0 = 6 × 0 is also divisible by 6.)½ mark

Check: n = 2: 8 − 2 = 6 = 6 × 1 ✓. n = 5: 125 − 5 = 120 = 4 × 5 × 6 = 6 × 20 ✓. n = 10: 1000 − 10 = 990 = 6 × 165 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Checking a few values (n = 2, 3, 4) and calling it proved. The question asks for reasons that work for every n; that needs the factorisation.
• Factorising n² − 1 as (n − 1)². It is a difference of squares: (n − 1)(n + 1).
• Saying “divisible by 2 and 3, so divisible by 6” without the reason. The step works because 2 and 3 have no common factor (12 is divisible by 4 and 6 but not by 24).

How this can come in the exam

Try one yourself

Show that n² + n is always even for every natural number n.

More questions like this

• Find the value of
• Consider any three consecutive square numbers. For example, 1, 4, and 9. Add the smallest and the largest squares. Thus, 1 + 9 = 10. Then subtract twice the middle square from this sum. This leads to 10 − (2 × 4) = 10 − 8 = 2. Now try the same process with another set of three consecutive square numbers. Say 9, 16, 25.
• Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.
• Think of numbers a and b where a and b do not represent lengths of line segments. What if a and b are negative numbers? Let us check for some negative numbers and see if this equation still works.
• What is the difference between an equation and an identity?

All Algebraic identities questions · All maths questions